Introductory Mathematical Analysis for Business Solution Manual – 2025 Step-by-Step Business Math Solutions

Chapterg0

Problemsg0.1 Problemsg0.2
,
1. False;g —13gisgnotgagrealgnumber. 1. False,gbecauseg0gdoesgnotghavegagreciprocal.

66g
2. True,gbecauseg— 2. True;g6.6 = sogthegreciprocalgofg6.6gis
10
2gandg7garegintegersgandg7g≠g0. 10
=g0.151515g.g.g..
66g
3. False,gbecauseg—3gisgnotgpositive.

0 3.g False;gthegnegativegofg7gisg—7gbecause
4.g False,gbecauseg =g 7g+g(—7)g=g0.
0 1g
.

5. False,gitgisgfairlygeasygtogseeg(althoughgthegdetai 4. True;g1(xg·gy)g=g(1g·gx)(1g·gy)
lsgaregnotgrelevantgforgagcoursegusinggthisgtext)
gthat 5. False;g—xg+gyg=gyg+g(—x)g=gyg—gx.
,
n,gforgnganginteger,gisgeithergangintegerg(ifgngi 6. True;g(xg+g2)(4)g =g (x)(4)g+g(2)(4)g =g 4xg+g8.
sga
perfectgsquare)gorgirrationalg(ifgngisgnotgagperfec x x 15 x 15 x 3
tgsquare).gThegperfectgsquaresgare
1,g4,g9 ,,1 6 , g25,g36,g.g.g.gandgsinceg3gisgnotgamon
g
these, 3gisgnotgrational. 7. false; +g3g=g gg+ g g g +g
g+ =g ≠g .
5 5 5 5 5
6. False;gwegcannotgdividegbyg0. gg gg
b abg
8. True,gbecausega =g .
7. , c c
False,gbecause 25g =g 5,g whichg isgag positive
integer.
,
, 14. False;g —13gisgnotgagrealgnumber.
8. True;g g2 isgangirrationalgrealgnumber.

9. False;gwegcannotgdividegbyg0.

10. False,gweghaveg3g<gvg<g4gsogthatgvgisgatgbest
grationalgandgnotganginteger.gItgcangbegshowngth
atgvgisgirrationalg(butgthegdetailsgaregnotgreleva
ntgforgourgpurposes).

11. True;gseegFigureg0.1gingthegtext.

12. False,gsincegthegintegerg0gisgneithergpositivegn
orgnegative.

13. True;gwegcangregardgagterminatinggsequencega
sgbeinggthegsamegasgthegsequencegobtainedgfro
mgitgbygappendingginfinitelygmanygzeros.
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l.

, 9. False;g2(xg·gy)g=g2xygwhile
(2x)g·g(2y)g =g (2g·g2)g·g(xg·gy)g =g 4xy.

10. True;gbygthegassociativegandgc
ommutativegproperties,gx(4y)g
=g(xg·g4)yg=g(4g·gx)yg=g4xy.

11. distributive

12. thegassociativegpropertygofgaddition

13. associative

14. definitiongofgdivisiongandgcommutativegproperty

15. commutativegandgdistributive

16. associative

17. thegdefinitiongofgdivision

18. commutative




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l.

,ISM:gIntroductoryg Mathematicalg Analysis Sectiong0.
3

19. distributivegandgcommutative 37.g (—1.6)(—0.5)g =g (1.6)(0.5)g =g 0.8

20. distributive 38. 19(—1)g=g(—1)19g=g—(1g·g19)g=g—19

g—1g g gg
21. 2x(yg—g7)g=g(2x)yg—g(2x)7g=g2xyg— 39. =g— —ga =ga
g(7)(2x)
1
=g2xyg—g(7g·g2)xg=g2xyg—g14x —g1a 1


22. x
gg
1
gg gg
1g
gg gg
1
gg 40. —(—6g+gx)g=g—(—6)g—gxg=g6g—gx
y z g= xyg zg=gxy z =gyx zg x wher
y g =g g
z e 41. —7(x)g=g—(7x)g=g—7x

weghavegused,gingorder,gthegdefinitiongofgdivisio 42. —3(ag—gb)g =g —3ag+g3bg=g 3(bg—ga)
n,gthegassociativitygofgmultiplication,gthegcomm
utativitygofgmultiplication,gandgthegdefinitiongof 43. —[—6g+g(—y)]g=g—(—6)g—g(—y)g=g6g+gy
gdivision
44. —3g 1g·g3g 1
—3g÷g(3a)g=g =g—g =g—g
23. (xg+gy)(2)g=g2(xg+gy)g=g2xg+g2y
24. a[bg+g(cg+gd)]g=ga[(bg+gc)g+gd 3a 3g·ga a
]
45. 9 g—9 9 9g·g1 1
=ga[dg+g(bg+gc)]g=g a[(dg+gb)g 27
+gc]
— )g=g =g =g =g
—27g 27g 9g·g3g 3
25. x[(2yg+g1)g+g3]g=gx[2yg+g(1g+g3)]g=gx[2yg g ÷g(—

+g4]
—ag a
=gx(2y)g+gx(4)g=g(xg·g2)yg+g4xg=g(2x)yg 46. (—a)g÷g(—b)g=g =g
+g4x
—b b
=g2xyg+g4x
gg gg
26. (1g+ga)(bg+gc)g=g 1(bg+gc)g+ga(bg+gc) 9
47. 3g+g(3—19)g=g3g+ 3 =g3g+g3g=g6
=g1(b)g+g1(c)g+ga(b)g+ga(c)g=gbg+gcg+gabg
+gac 48. 3[—2(3)g+g6(2)]g=g3[—6g+g12]g=g3[6]g=g18
27. (xg—gyg+gz)wg=g((xg+g(—y))g+gz)w
49. (—a)(—b)(—1)g=gab(—1)g=g—ab
=g(xg+g(—y))wg+gzwg=gxwg+g(—y)wg+gzw
=gxwg+g(—yw)g+gzwg=gxwg—gywg+gzw 50. (—12)(—12)g =g (12)(12)g =g 144

28. —2g+g(—4)g=g—6 51. X(1)g=gX

29. —ag+gbg=gbg—ga 52. —71(xg—g2)g=g—71xg+g142g=g142g—g71x

30. 6g+g(—4)g=g2 53. 4(5g+gx)g=g4(5)g+g4(x)g=g20g+g4x

31. 7g—g2g=g5 54. —(xg—gy)g=g—xg+gyg=gyg—gx
3
3 3g g 2 6
1
32. = 1 = 1g·g 1 = 1 = 6 55. 0(—x)g =g 0
2—1 2
g
33.g —5g—g(—13)g=g— 34. —(—a)g+g(—b)g=gag—gb
= =
5g+g13g=g8
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l.

, Chapterg0:g ReviewgofgAlgebra ISM:gIntroductoryg Mathematicalg Analysis



35. (—2)(9)g=g—(2g·g9)g=g—18 g1 8g·g1 g8
56. 8g g
11 11 11
X
57. =X
1
36. 7 9 7g 9 63 14xg 2g·g7g·gxg 2x
58. =g =g
(—g )g=g— )g=g 21y 3g·g7g·gy 3y
(g · —




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l.